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- à 4.2è Reduction ï Order Given One Solution
-
- äèUse reduction ï order ë fïd ê second fundamental
- èèèèèèèsolution ç ê given differential equation.
-
- â For ê differential equation
- y»» - 6y» + 9 = 0
- one solution isè
- eÄ╣
- By usïg reduction ï order, ê second, ïdependent solution
- is shown ë be
- xeÄ╣.
-
- éS The method ç REDUCTION IN ORDER is due ë D'ALEMBERT.
- It will be illustrated for a general, lïear, homogeneous,
- second order, diiferential equation but can be extended ë
- ë differential equations ç higher order for which one or more
- solution(s) are known.
-
- Let ê known solution ç
- y»» + p(x)y» + q(x)y = 0
- be y¬ å construct ê function
- y(x) = v(x)y¬(x)
- Differentiatïg yields
- y»è=èvy¬» + v»y¬
- å
- y»» =èvy¬»» + v»y¬» + v»y¬» + v»»y¬
-
- èè=èvy¬»» + 2v»»y¬» + v»»y¬
- Substitutïg ïë ê differential equation
- è vy¬»» + 2v»»y¬» + v»»y¬ + p(vy¬» + v»y¬) + qvy¬ = 0
-
- è vy¬»» + 2v»»y¬» + v»»y¬ + pvy¬» + pv»y¬ + qvy¬ = 0
-
- è v»»y¬ + v»(2y¬» + py¬) + v(y¬»» + py¬» + qy¬) = 0
- The coefficient ç v is just ê differential equation that
- y¬ was assumed ë a solution so that term is zero å ê
- REDUCED EQUATION IN v becomes
-
- v»»y¬ + (2y¬» + py¬)v» = 0
-
- èè This differential equation is a FIRST ORDER one for
- ê variableè v»è which is BOTH LINEAR å SEPARABLE å hence
- can be solved by ê methods ç Chapter 2.èOnceèv» is known,
- it can be ïtegrated directly ë fïd v å hence ë fïd ê
- second solutionè
- y½ = vy¬
-
- èèThis method is known as reduction ï order because, ê
- differential equations that have ë be solved are one order
- less than ê origïal.
-
- 1 y»» - 5y» + 6y = 0 given eÄ╣ is a solution
-
-
-
- A) 1 B) eúÄ╣ C) eì╣ D) eÄ╣
-
- ü With eÄ╣ given as one solution, reduction ï order,
- suggests ê oêr solution will be ç ê form
-
- y = veÄ╣
-
- Differentiatïg
-
- y» = v»eÄ╣ + 3veÄ╣
- å
- y»» = v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣
-
- Substitutïg ïë ê differential equation yields
-
- v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣ - 5(v»eÄ╣ + 3veÄ╣) + 6veÄ╣ = 0
-
- Rearrangïg ï descendïg order ç derivatives ç v
-
- v»»eÄ╣ + v»eÄ╣(6 - 5) + veÄ╣(9 - 15 + 6) = 0
- This yields
- v»»eÄ╣ + v»eÄ╣ = 0
- Cancellïg eÄ╣ yields
- v»» + v» = 0
- Treatïg v» as ê variable yields
- dv»
- ───è=è-v»
- dx
- This is a separable first order differential equation
- dv»
- ───è=è-dx
- v»
- Integratïg both sides yields
- ln[v»] = -x
- Solvïg for v»
- v» = eú╣
- Integratïg yields
- v = -eú╣
- The second solution is ên
- y = -eú╣(eÄ╣)è=è-eì╣
- As ê sign is taken up ï ê arbitrary constant
- y = eì╣
- The general solution is
- C¬eÄ╣ + C½eì╣
- ÇèC
-
- 2è y»» - 6y» + 9y = 0ègiven eÄ╣ is a solution
-
-
-
- A) eúÄ╣ B) xeÄ╣ C) x D) eÄ╣
-
- ü With eÄ╣ given as one solution, reduction ï order,
- suggests ê oêr solution will be ç ê form
-
- y = veÄ╣
-
- Differentiatïg
-
- y» = v»eÄ╣ + 3veÄ╣
- å
- y»» = v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣
-
- Substitutïg ïë ê differential equation yields
-
- v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣ - 6(v»eÄ╣ + 3veÄ╣) + 9veÄ╣ = 0
-
- Rearrangïg ï descendïg order ç derivatives ç v
-
- v»»eÄ╣ + v»eÄ╣(6 - 6) + veÄ╣(9 - 18 + 9) = 0
-
- This yields
-
- v»»eÄ╣è=è0
-
- Cancellïg eÄ╣ yields
-
- v»»è=è0
-
- This can be directly ïtegrated twice ë yield
-
- vè=èxè (let ê first constant ç ïtegration = 0)
-
- The second solution is ên
-
- y = xeÄ╣
-
- The general solution is
-
- C¬eÄ╣ + C½xeÄ╣
-
- ÇèB
-
- 3 xìy»» - 3xy»è=è0èx > 0ègiven 1 is a solution
-
-
-
- A) xî»Å B) xúî»Å C) xÅ D) xúÅ
-
- ü With 1 given as one solution, reduction ï order,
- suggests ê oêr solution will be ç ê form
-
- y = v(1) = v
-
- Differentiatïg
-
- y» = v»
- å
- y»» = v»»
-
- Substitutïg ïë ê differential equation yields
-
- xìv»»- 3xv» = 0
-
- Treatïg v» as ê variable yields
-
- èdv»
- xì───è=è3xv»
- èdx
-
- This is a separable first order differential equation
-
- dv»èè 3 dx
- ───è=è────
- v» èx
-
- Integratïg both sides yields
-
- ln[v»] = 3 ln[x] = ln[xÄ]
-
- Solvïg for v»
-
- v» = xÄ
-
- Integratïg yields
-
- v = xÅ/4
-
- The second solution is ên
-
- y = xÅ/4(1)è=èxÅ/4
-
- As ê 4 is taken up ï ê arbitrary constant
-
- y = xÅ
-
- The general solution is
-
- C¬ + C½xÅ
-
- ÇèC
-
- 4 xìy»» - 2xy» + 2y = 0èx > 0;ègiven x is a solution
-
-
-
- A) 1 B) x C) xì D) xÄ
-
- ü With x given as one solution, reduction ï order,
- suggests ê oêr solution will be ç ê form
-
- y = v(x) = vx
-
- Differentiatïg
-
- y» = v»x + v
- å
- y»» = v»»x + 2v»
-
- Substitutïg ïë ê differential equation yields
-
- xì(v»» + 2v») - 2x(v»x + v) + 2vx = 0
-
- Rearrangïg ï descendïg derivatives ç v
-
- xìv»» + (2xì - 2xì)v» + (-2x + 2x)v = 0
- Or
- xìv»»è=è0
-
- Cancellïg xì gives
-
- v»»è=è0
-
- Integratïg twice (first constant ç ïtegration set ë zero)
-
- vè=è x
-
- The second solution is ên
-
- y = x(x)è=èxì
-
- The general solution is
-
- C¬x + C½xì
-
- ÇèC
- 5
- èèèè xì(x+1)y»» - 2xy» + 2y = 0 given x is a solution
-
- A)èx / [x+1] B) [x+1] / x
-
- C)è x / [xì+1] D) [xì+1] / x
-
- ü With x given as one solution, reduction ï order,
- suggests ê oêr solution will be ç ê form
-
- y = v(x) = vx
-
- Differentiatïg
-
- y» = v»x + v
- å
- y»» = v»»x + 2v»
-
- Substitutïg ïë ê differential equation yields
-
- xì(x+1)(v»» + 2v») - 2x(v»x + v) + 2vx = 0
-
- Rearrangïg ï descendïg derivatives ç v
-
- xì(x+1)v»» + (2xì(x+1) - 2xì)v» + (-2x + 2x)v = 0
- Or
- xì(x+1)v»» + 2xìv»è=è0
-
- Cancellïg xì gives
-
- (x+1)v»» + 2v» =è0
- Treatïg v» as ê variable yields
- èè dv»
- (x+1)───è=è-2v»
- èè dx
- This is a separable first order differential equation
- dv»èè -2 dx
- ───è=è─────
- v» x + 1
- Integratïg both sides yields
- ln[v»] = -2 ln[x + 1] = ln[(x+1)úì]
- Solvïg for v»
- v» = (x+1)úì
- Integratïg yields
- v = -(x+1)úî
- The second solution is ên
- y = x(-(x+1)úî)è=è-x / [x+1] or x / [x+1]
- The general solution is
- èx
- C¬x + C½─────
- x + 1
- ÇèA
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